Answer:
We need 10.14 grams of sodium bromide to make a 0.730 M solution
Explanation:
Step 1: Data given
Molarity of the sodium bromide (NaBr) = 0.730 M
Volume of the sodium bromide solution = 135 mL = 0.135 L
Molar mass sodium bromide (NaBr) = 102.89 g/mol
Step 2: Calculate moles NaBr
Moles NaBr = Molarity NaBr * volume NaBr
Moles NaBr = 0.730 M * 0.135 L
Moles NaBr = 0.09855 moles
Step 3: Calculate mass of NaBr
Mass NaBr = 0.09855 moles * 102.89 g/mol
Mass NaBr = 10.14 grams
We need 10.14 grams of sodium bromide to make a 0.730 M solution
Answer:
Chromosomes are structures within cells that contain a person's genes.
Explanation:
Genes are segments of deoxyribonucleic acid (DNA) that contain the code for a specific protein that functions in one or more types of cells in the body. Chromosomes are structures within cells that contain a person's genes. Genes are contained in chromosomes, which are in the cell nucleus.
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Answer:
- Initial: forward rate > reverse rate
- Equilibrium: forward rate = reverse rate
Explanation:
2NO₂(g) → N₂O₄(g) Kc=4.7
The definition of <em>equilibrium</em> is when the forward rate and the reverse rate are <em>equal</em>.
Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.
